Well... depends if they were walking around and meeting each other anyway...
Here's why it matters:
In the case of only 1 blue-eyed person, if he's not aware that at least 1 is blue, he may assume that everyone's brown, including himself, so he'll never leave.
If there were 2, they each see 1, but they can't logically deduce anything from that, because 1 will never leave, if he's not made aware that there's at least 1. ALTHOUGH, REALIZE THAT EVERYONE KNOWS THERE ARE AT LEAST ONE, CUZ THEY ALL SEE AT LEAST ONE.
Here, the fact that everyone knows is not enough, but what if someone announces that at least one is blue? Now, they all know that everyone knows. So if person #2 knows that person #1 also knows, they can now use the reasoning in the solution.
If there were 3, they each see 2, so they alreadyknow that everyone knows at least one is blue, but here the announcement will tell them that everyone knows that everyone knows that everyone knows.
and so on and so forth till 40.
Here's why it matters:
In the case of only 1 blue-eyed person, if he's not aware that at least 1 is blue, he may assume that everyone's brown, including himself, so he'll never leave.
If there were 2, they each see 1, but they can't logically deduce anything from that, because 1 will never leave, if he's not made aware that there's at least 1. ALTHOUGH, REALIZE THAT EVERYONE KNOWS THERE ARE AT LEAST ONE, CUZ THEY ALL SEE AT LEAST ONE.
Here, the fact that everyone knows is not enough, but what if someone announces that at least one is blue? Now, they all know that everyone knows. So if person #2 knows that person #1 also knows, they can now use the reasoning in the solution.
If there were 3, they each see 2, so they alreadyknow that everyone knows at least one is blue, but here the announcement will tell them that everyone knows that everyone knows that everyone knows.
and so on and so forth till 40.
I agree that if there was only 1 blue-eyed person on they islnad, they would live there happily ever after if they were not told that there is at least one.
But the information becomes redundant as soon as as you see plenty blue-eyed people running around...
That's why I said I suppose this information is a rest of a different permutation of the same puzzle...
I agree that if there was only 1 blue-eyed person on they islnad, they would live there happily ever after if they were not told that there is at least one.
But the information becomes redundant as soon as as you see plenty blue-eyed people running around...
That's why I said I suppose this information is a rest of a different permutation of the same puzzle...
Im glad you’re saying this. I also only find it necessary until there are 3 blue eyed guys. But had no heart to make her explain things again. Once there are 3 then everyone knows that everyone knows.
I agree that if there was only 1 blue-eyed person on they islnad, they would live there happily ever after if they were not told that there is at least one.
But the information becomes redundant as soon as as you see plenty blue-eyed people running around...
That's why I said I suppose this information is a rest of a different permutation of the same puzzle...
It is not, because were it not publicly made aware that at least one is blue, then the whole logic cannot be deduced.
Were there 4 blue-eyed people, how can they tell by day #3 if they themselves are blue? These 3 people won't know to leave, because seeing 2, they are waiting till day 2, but these 2 ppl won't know to leave either, since each sees one who might not be aware theres at least one....
Im glad you’re saying this. I also only find it necessary until there are 3 blue eyed guys. But had no heart to make her explain things again. Once there are 3 then everyone knows that everyone knows.
perhaps it's the foundation to set this whole recurrence in motion...
It is not, because were it not publicly made aware that at least one is blue, then the whole logic cannot be deduced.
Were there 4 blue-eyed people, how can they tell by day #3 if they themselves are blue? These 3 people won't know to leave, because seeing 2, they are waiting till day 2, but these 2 ppl won't know to leave either, since each sees one who might not be aware theres at least one....
Oh ok. I got this quicker than the solution
To get the ball rolling the leader must announce it.
How can you get to the number 21 using 1, 5, 6 and 7? You may use each digit only once and standard arithmetic operations (addition, subtraction, multiplication and division) as many as you want. You may also use parentheses. Any other symbols are not allowed I.e. decimals.
You CANNOT combine two numbers. For example, 15 + 6 = 21...you can't combine 1 and 5 to make 15.
If you're just saying you got it, cuz you don't want to make me repeat it, that's also fine
Just noticed this. Nope I will never say I get it if I don’t. If anything I’ll say never mind skip no patience left... I think my previous post proves that I really got it.
Just noticed this. Nope I will never say I get it if I don’t. If anything I’ll say never mind skip no patience left... I think my previous post proves that I really got it.
I know you got it. But does everyone know that everyone knows that everyone knows that you got it?
Tamari I’m sorry but I still don’t get why they had to mention there is at least one. When it’s basically a fact to everyone. Even if by solving The riddle you go back to the core and start trying with one, it still should not matter for now.
In trying to solve the riddle using 4 people....they never need to be toLd that there is at least one either...since each of the 4 people can see at least 3 then they only need to go back as far as thinking if there are only 3 blue eyed pple then after day 3 they will leave ( because they would assume there’s only 2 until they see themselves. But never lower than that....
Tamari I’m sorry but I still don’t get why they had to mention there is at least one. When it’s basically a fact to everyone. Even if by solving The riddle you go back to the core and start trying with one, it still should not matter for now.
In trying to solve the riddle using 4 people....they never need to be toLd that there is at least one either...since each of the 4 people can see at least 3 then they only need to go back as far as thinking if there are only 3 blue eyed pple then after day 3 they will leave ( because they would assume there’s only 2 until they see themselves. But never lower than that....
Just to start the count. On day one it’s a definite 1 makes it a definite 2 on day 2 till you get a definite 40 on day 40. Otherwise on day 40 you might still think there are 39 and never know for sure otherwise.
